April 30th, 2017, 02:39 AM  #1 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 154 Thanks: 1  Summation Problem
Hello Guys! Can someone help me with this! How many pair of positive integers of $(m,n)$ are there satisfying $\sum_{i=1}^{n} i! = m!$ I guess it has only one pair of solution : $(m,n) =(1,1)$ Then starting and ending value of Summation should be equal as I start at $i=1$ and we need $n$ to be 1. Will it make sense or Is it wrong to have both values same ? Please let me know Thank you 
April 30th, 2017, 04:00 AM  #2 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,277 Thanks: 204 
For $ \ \ \ \ n > 1 \ \ \ \ $ , $ \ \ \ \ n < m \ \ \ \ $ otherwise equality has no chance. Is this obvious to you? Suppose $ \ \ \ \ n = m  1$ We need to determine if $ \ \ \ \ \sum_{i=1}^{m1}i! \ \ \ \ $ can catch up to $ \ \ \ \ m!$ Let's see how the desired equality would look $ 1! + 2! + 3! + ... + (m1)! = 1 \times 2 \times 3 \times ... \times (m1) \times m $ $ 1! + 2! + 3! + ... + (m1)! = (m1)! \times m $ $ 1! + 2! + ... + (m1)! = (m1)! + (m1)! + ... + (m1)! $ The left hand side has $ \ \ \ \ m1 \ \ \ \ $ sums , the right hand side has $ \ \ \ \ m \ \ \ \ $ sums. Moreover , every sum on the left except the last sum is smaller than any sum on the right. So no chance to catch up Last edited by agentredlum; April 30th, 2017 at 04:06 AM. 
April 30th, 2017, 04:29 AM  #3 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,277 Thanks: 204 
You can have both values the same $ \sum_{i = 1}^{1} f(i) = f(1) $ 
April 30th, 2017, 04:35 AM  #4 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 154 Thanks: 1  

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